Been using this workbook the past 24 hours and I'm half way thru it and it's really helping me breeze through subnetting; I learned subnetting before I switched gears onto the VCP5-DCV track and lost touch with it but from my superb notes and this work book that I noted in my 5-star massive 6 subject notebook to do in-case I forgot (and see I knew I would) I'm picking it up again.. Right now I can do one in about 45 seconds; there is still 45 pages left of worksheets so maybe I'll get better at it.

See I can subnet Class B and C fine in binary but once I get to Class A remembering those large A** numbers is daunting to say the least. Curious if the CCENT/CCNA will have Class A's on it or if it's really just tons and tons of practice to just remember them?

The past 3 weeks I've been doing on average 20 or so subnetting practices a day (when I have down-time at work) between my workbook above and subnetting.net; and watch every subnetting video on youtube.

I don't expect you will need anything past a /22.
If you do then you can revert to the math quickly on the whiteboard they give you. If you are familiar with binary up to like 4096 (/20) I don't think you will have any problems.

I don't expect you will need anything past a /22.
If you do then you can revert to the math quickly on the whiteboard they give you. If you are familiar with binary up to like 4096 (/20) I don't think you will have any problems.

Thanks Jon, I can do Class A but with a calculator. I've been just doing Class B and C's and really honing them. I mean If I had to do a Class A I could do it but with long math, just it's tedious to remember those numbers and I suck with multiplication. I haven't used math is like 6 years since college....

I understood how to do the example originally posted and went through most of the examples posted by others , but i could not find answer to this type of question and this is the tricky ones that I am not able to get my head aroundWhat is the broadcast address of the network 172.28.74.192 255.255.255.192?Answer: 172.28.74.255

What i did was took the block size as 64

and my subnets were

172.28.0.0
172.28.64.0
172.28.128.0
so going by this logic the address specified in the question belongs to 172.28.64.0 and the broadcast address will be 172.28.127.255 .

But the answer says its 172.28.74.255 where the 3rd octet remained unchanged. But if it is a Class B Address (starting with 172) then should the 3rd octet change ?

Thank you for your sheet but am not clear on how your table translates. I know that by default that a CLass B would be a /16. With getting 300 hosts this would now become a /25 ( Am thinkinh... not sure ) BUT how do you take /25 to make up the netmask- 255.255.254.0

Edit. No answers required any more. Worked it out, albeit using another method. Throwing into questions now to get it all nailed down.

Thank you for your sheet but am not clear on how your table translates. I know that by default that a CLass B would be a /16. With getting 300 hosts this would now become a /25 ( Am thinkinh... not sure ) BUT how do you take /25 to make up the netmask- 255.255.254.0

Edit. No answers required any more. Worked it out, albeit using another method. Throwing into questions now to get it all nailed down.

Thanks

For anyone reading this, where he went wrong is that you add subnets to the subnet mask or subtract host from /32, so 32 - 9 = /23 OR /16 + 7 bits of subnets (128 subnets) = /23

Originally Posted by Deathmage

making sure I got this Class A address right:

number of needed hosts: 29
Network Address: 23.0.0.0

so.....

Address Class: A
Default Mask: 255.0.0.0
Custom Mask: 255.255.255.224
Total number of Subnets: 524,288
Total number of Host addresses: 32
Total usable addresses: 30
Number of bits borrowed: 19

Does that look right; I can remember up to 65,538 but need a calculator to go any higher. Really hope the exam doesn't have anything higher than 65,538.

Again for anyone reading... I have a few power of two "anchor points" if you will, and think of them as MegaBytes for thinking.

So
2^3 = 8MB
2^6 = 64MB2^10 = 1GB (1024)

For 2^ 10, 11, 12, 13, it's really easy, it's 1024, 1024*2, *4, or *8

For mental math, this is how I do the bigger numbers, it seems to work best for my mind, though I'll switch it up with different techniques at times. Useful for interviews I guess? I tend to do mental over calculator by habit

For 14 to 23, you do -10 to the exponent. Let's call it x. We want to multiply it by 1024 (2^10). x*1000 + 2x*10 + 2*2x

Hello, can someone done this task for me?An ISP is granted a block of addresses starting with 158.78.0.0/16. The ISP wants to distributethese blocks to 3200 customers as follows.a. The first group has 400 medium-size businesses; each needs 64 addresses.b. The second group has 600 small businesses; each needs 16 addresses.c. The third group has 80 households; each needs 256 addresses.Design the subblocks and give the slash notation for each subblock. Find out how manyaddresses are still available after these allocations

Hi all, can you please check if below answers are correct, as I have been given below for next coming interview questions.
Thanks in advance for the assistance.

1. Enter the last valid host on the network 192.168.92.0 255.255.255.224?
Increment 32, so the last valid host is 192.168.92.30

2. What is the maximum number of valid subnets one will have from the network 10.29.252.144/13? Assume this is a class A address.
2^5 = 32 maximum number of valid subnets, since borrowing 5 bits

3. What is the most efficent subnet if you need 110686 usable hosts on a subnet? Present your answer as a subnet mask?
2^17 - 2 = 131070. So the subnet mask will be 255.254.0.0, as we need to have 17 zero bits for the host.

4. What is the broadcast address of the network 10.121.108.13/14?
Network ID: 10.120.0.0, 10.124.0.0
So, the broadcast is 10.123.255.255

5.What is the maximum number of valid hosts one will have from the network 192.168.124.0 255.255.255.224?
5 bits zero for the host, so the answer is 30

6. What is the number of network IDs in a Class C network? 3, because of 255.255.255.0

Lordflasheart, I don't know if you came up with this on your own? But it's amazing. I'm not as advanced as some (I'm working on my net+ cert) but this is very helpful and getting the basics down. Thanks, I really appreciate it!

Question: What is the first valid host on the subnetwork that the node 172.18.186.50/23 belongs to?

I'm confused a little - you've asked the question that is clearly copied and from a website, on a forum that is about technology. AND on a subnetting thread that TEACHES you how to subnet. This one is easy to answer. How about you have a little stab at it. If you get it wrong, we can help you.

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